Let $X$ be a connected $CW$-complex, such $\pi_1(X)$ is torsion-free and $H_k(X,\mathbb Z) = 0$ for all $k \geq N$ and some $N \in \mathbb N$. Then

$(1)$ Does it follow that $X$ is homotopy-equivalent to its $N$-skeleton, i.e $X \simeq X^{(N)}$ ?

$(2)$ If $(1)$ is false, does it follow that $X \simeq X^{(k)}$ for some $k \in \mathbb N$ ?

$(3)$ If $(2)$ is also false, at the very least, does it follow that $X \simeq Y$ for some finite-dimensional $CW$-complex $Y$ ?

It is clear to me that the question has a negative answer when $H_k$ is replaced by $\pi_k$ (take a $CW$-model of $BG$ for a group $G$ containing elements of finite order). I am also very certain that this question has been asked before and I just couldn't find the corresponding thread, so I have no problem with this question being marked as duplicate.

Edit: One could then possibly find a finitely dominated $CW$-complex $X$, with $\pi_1(X)$ not torsion-free, such that its Wall finiteness obstruction element $w(X) \in \widetilde{K_0}(\mathbb Z[\pi_1(X)])$ is non-trivial. Any finitely dominated $CW$-complex has finite homological-dimension (so it satisfies the assumptions), but if its finiteness obstruction doesn't vanish, it doesn't have the homotopy type of a finite complex.

Conversely, if $X$ is finitely-dominated and $w(X) = 0$, then it has the homotopy type of a finite-dimensional complex.

In particular, this must be the case if $\widetilde{K_0}(\mathbb Z[\pi_1(X)]) =0$. It is conjectured that $\widetilde{K_0}(\mathbb Z[G]) = 0$ if $G$ is torsion-free, hence the assumption in the statement of the question.

Edit 2: Mark has given an example of a $CW$-complex $X$ with finite homological dimension, but infinite cohomological dimension (and thus, has given a negative answer to my original questions) . It would be still interesting to know if there exist counter-examples $X$, such that for any $\pi_1(X)$-module $M$, both the homology $H_*(X,M)$ and the cohomology $H^*(X,M)$ have groups only in finitely many dimensions.

  • $\begingroup$ Could you explain your last remark a little more? If G is finite, say Z/2, then BG is RP-inf and the BG and the cohomology is periodic. Did you have a specific theorem in mind? $\endgroup$ – Asvin Sep 11 '18 at 13:53
  • $\begingroup$ Yes, the cohomolgy is periodic, so in particular, there is cohomology in arbitrary high dimensions, which implies that $BG$ cannot be homotopy equivalent to a finite-dimensional $CW$-complex. $\endgroup$ – H1ghfiv3 Sep 11 '18 at 13:55
  • $\begingroup$ Ah sorry, I somehow missed that you had switched to the homotopy groups despite reading it clearly. I am just going to blame the mental fog that comes with waking up :( $\endgroup$ – Asvin Sep 11 '18 at 13:57
  • $\begingroup$ I'm not sure I understand what the recent edit is asking, now. Does a finite product of acyclic groups give a counter-example to (1) (but not (2) and (3))? $\endgroup$ – Mark Grant Sep 11 '18 at 16:18
  • $\begingroup$ I think if a complex $X$ has finite cohomological dimension $n$, then it can be shown using obstruction theory that $X$ is h.e to a complex of dimension $\max(n,3)$. And with arbitrary coefficients, I think homological dimension is either equal to, or one less than, cohomological dimension. Do these observations answer your updated question? $\endgroup$ – Mark Grant Sep 11 '18 at 16:18

You can take $X=BG$ where $G$ is a torsion-free, acyclic group of infinite cohomological dimension. Acyclic means $H_k(BG;\mathbb{Z})=0$ for $k>0$, and infinite cohomological dimension implies infinite geometric dimension, so $BG$ is not homotopy equivalent to any finite dimensional CW complex $Y$. Such a group therefore gives a (rather extreme) counter-example to (1), (2) and (3).

I believe an infinite direct product of copies of Higman's group $H$ gives such a $G$. It is obviously torsion-free (since $H$ is), and is acyclic since homology commutes with direct limits and a finite product of acyclic groups is acyclic. Since $H$ is of type $FP_\infty$ (iirc, the presentation complex is a model for $BH$), we can apply Benjamin Steinberg's answer here to conclude that $G$ has infinite cohomological dimension.

  • $\begingroup$ I see, I wasn't aware that there are spaces with finite homological dimension, but infinite cohomological dimension! I wonder if one would still find counter-examples if both dimensions are assumed to be finite.... $\endgroup$ – H1ghfiv3 Sep 11 '18 at 14:47
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    $\begingroup$ Since you are only taking homology with trivial coefficients, and I'm allowed to take cohomology with arbitrary twisted coefficients, there are going to be lots of such examples. The stronger (and more common) notion of homological dimension $\le k$ asks that $H_i(X;M)=0$ for all $i>k$ and all $\pi_1(X)$-modules $M$. Then I think homological and cohomological dimensions are more closely related (differ by at most $1$?). $\endgroup$ – Mark Grant Sep 11 '18 at 14:57
  • $\begingroup$ Okay, thank you very much for pointing this out and providing me with this illuminating answer. Will you allow me to change the assumptions of my original question accordingly (I will, of course, point out that you have solved the original question) ? $\endgroup$ – H1ghfiv3 Sep 11 '18 at 14:59

For an elementary example, give $B^{n+1}$ the CW structure with a $0$-cell, a boundary $n$-cell, and an $n+1$-cell, then take $X=\bigvee_nB^{n+1}$. Then $X$ is contractible but $X^{(n)}$ is homotopy equivalent to $S^n$ (from the $B^{n+1}$ term) so $X^{(n)}$ is never homotopy equivalent to $X$.

It would be better to look for $Y$ with $X^{(n)}\subseteq Y\subseteq X^{(n+1)}$ such that the inclusion $Y\to X$ is a homotopy equivalence. I suspect that that is not possible either, although it would be a bit more delicate to provide an example. Better still, we can ask whether there is a CW complex $Y$ obtained from $X^{(n)}$ by attaching some $n+1$-cells, such that the inclusion $X^{(n)}\to X$ extends over $Y$, and the resulting map $Y\to X$ is a homotopy equivalence. The attaching maps for $Y$ will lie in the relative homotopy group $\pi_{n+1}(X,X^{(n)})$, which maps to $H_{n+1}(X,X^{(n)})$. If $X$ is simply connected then you can use the relative Hurewicz theorem and I think it works out that you can always find such a $Y$ provided that $H_{n+1}(X)$ is free abelian (perhaps zero) and $H_k(X)=0$ for $k\leq n$. If $X$ is not simply connected then you need to worry about the Wall finiteness obstruction.

  • $\begingroup$ Thanks. Just as you posted this answer, I realized, just like you did, that statement $(1)$ of my question must be clearly false. Moreover, just as you were posting the answer, I added a paragraph on Wall's Finiteness Obstruction. $\endgroup$ – H1ghfiv3 Sep 11 '18 at 14:30

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